Question

我花了1.5个小时的时间来完成这件事，而我却无法让它发挥作用。在它开始抛出这些奇怪的语法错误之前，它会继续说“猜到低，再试一次”。它甚至没有在我遇到这些错误的shell中运行。

以下是代码：

import random
gend = random.randint(1,100) #make random number
def compare(gend,guess): #start function

    guessnum = 0 #set guessnumber to zero for later printing

    if gend < guess: #compare gend number against guessed number
        int(raw_input("Too high. Try again. ")) #request another number for user
        guessnum = guessnum + 1 #add 1 to guessnumber, for later printing
    elif gend > guess: #compare gend number against guessed number
        int(raw_input("Too low. Try again. "))#request another number for user
        guessnum = guessnum + 1 #compare gend number against guessed number
    elif gend == guess: #if guessed number=gend, do this
            keepalive = 'much spook' #stop the while loop
            print "Congratulations! You got it in %d guess." % (guessnum)#print da amount of guesses

keepalive = 'rekt' #keepalive substitute

print 'Time to play a guessing game' #no exp needed

guess = int(raw_input("Enter a number between 1 and 100: ") #start off the game

while keepalive == 'rekt': #while loop that does all the work
    compare(gend,guess) #funcion werk

raw_input("Any key to exit") #keep console open so it doesnt autoclose -.-

我还要感谢你们的帮助，可以说是大多数语言的最佳学习资源。

Answer 1

您在语法错误上方的行上缺少右括号。

guess = int(raw_input("Enter a number between 1 and 100: ") #start off the game
                                                           ^

Answer 2

语法错误是由之前的行引起的。 Python只能告诉你注意到某些东西的关闭，这不一定是事情真正走下坡路的地方。 guess = int(raw_input("Enter a number between 1 and 100: ")中缺少一个结束副本。

它一直说“太低，再试一次”是由于两个错误：

当你要求另一个号码时，你不会分配给guess（你读了一个号码，解析它，然后扔掉它），所以guess永远不会改变。
在函数内部，keepalive = 'much spook'不会更改在循环外使用的keepalive。有几种方法可以解决这个问题，但哪一种方法是合适的取决于您之前的经验和您正在学习的材料 - 请参考它以了解是否提到了类似的内容。

猜测游戏不起作用，也会抛出奇怪的语法错误

2 个答案: